Question

Solve the quadratic equation 12x2−12x+4=−3x
. Which of the following expresses its solutions in the form a±bi
?(1 point)
Responses

38+111√24i
Start Fraction 3 over 8 End Fraction plus Start Fraction Start Root 111 End Root over 24 End Fraction i

−38±111√24i
negative Start Fraction 3 over 8 End Fraction plus or minus Start Fraction Start Root 111 End Root over 24 End Fraction i

38±111√24i
Start Fraction 3 over 8 End Fraction plus or minus Start Fraction Start Root 111 End Root over 24 End Fraction i

38±273√24i

Answer 1

To solve the quadratic equation \(12x^2 - 12x + 4 = -3x\), we first rearrange the equation:

\[
12x^2 - 12x + 4 + 3x = 0
\]
\[
12x^2 - 9x + 4 = 0
\]

Now we can use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 12\), \(b = -9\), and \(c = 4\).

First, we need to calculate the discriminant:

\[
b^2 - 4ac = (-9)^2 - 4(12)(4)
\]
\[
= 81 - 192
\]
\[
= -111
\]

Since the discriminant is negative, the solutions will be complex. Next, we substitute into the quadratic formula:

\[
x = \frac{-(-9) \pm \sqrt{-111}}{2(12)}
\]
\[
= \frac{9 \pm \sqrt{-111}}{24}
\]

We can express \(\sqrt{-111}\) as \(i\sqrt{111}\):

\[
x = \frac{9 \pm i\sqrt{111}}{24}
\]

Now we can split this into real and imaginary parts:

\[
x = \frac{9}{24} \pm \frac{\sqrt{111}}{24}i
\]
\[
= \frac{3}{8} \pm \frac{\sqrt{111}}{24}i
\]

Thus, the solutions in the required form \(a \pm bi\) are:

\[
\frac{3}{8} \pm \frac{\sqrt{111}}{24}i
\]

Now, looking at your options, we see that this corresponds to:

**Start Fraction 3 over 8 End Fraction plus or minus Start Fraction Start Root 111 End Root over 24 End Fraction i**.